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Charge densities and delta functions

  1. Sep 16, 2016 #1
    Note from mentor: this thread was originally posted in a non-homework forum, therefore it lacks the homework template.

    I was wondering if the electric charge density ##\rho({\bf{r}})## of a point charge ##q## at position ##{\bf{r}}_{0}## is given by ##\rho({\bf{r}})=q\delta^{(3)}({\bf{r}}-{\bf{r}}_{0})##.

    If so, is this because ##\int d^{3}{\bf{r}}\ \rho({\bf{r}}) = \int d^{3}{\bf{r}}\ q\delta^{(3)}({\bf{r}}-{\bf{r}}_{0}) = q##?
     
  2. jcsd
  3. Sep 16, 2016 #2

    vanhees71

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    If it's a point charge at rest, yes. Note that ##\delta^{(3)}(\vec{r}-\vec{r}_0)## is NOT a function BUT a distribution (generalized function)!
     
  4. Sep 16, 2016 #3
    Let's say that now we have a point charge ##-q## at the origin and a point charge ##+q## at position ##\bf{r}_{0}##. Is the charge density of the dipole given by ##\rho({\bf{r}})=-q\delta^{(3)}({\bf{r}})+q\delta^{(3)}({\bf{r}-\bf{r}_{0}})##?
     
  5. Sep 16, 2016 #4

    vanhees71

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    Yes, but usually a dipole is defined as the approximation that ##r_0## is very small in the sense of the multipole expansion.
     
  6. Sep 16, 2016 #5
    Thanks.

    I'm also trying to find the charge density of a uniform infinitesimally thin spherical shell of radius ##R## and total charge ##Q## centred at the origin.

    This is my attempt:

    ##\int\ \rho({\bf{r}})\ dV = Q##

    ##\lim\limits_{\epsilon\to 0}\ \int\limits_{R-\epsilon}^{R+\epsilon}\ \rho({\bf{r}})\ (\text{sin}\theta\ d\theta\ d\phi)\ (r^{2}dr) = Q##

    ##\lim\limits_{\epsilon\to 0}\ \int\limits_{R-\epsilon}^{R+\epsilon}\ \rho({\bf{r}})\ (r^{2}dr) = \frac{Q}{4\pi}##

    Any hints on what to do next?
     
  7. Sep 21, 2016 #6

    rude man

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    How about assuming ρ(r) = f(r) δ(r - r0) in your last integral
    because you know ρ will have to have an infinity at r0 to cancel out the dr in Q = ∫ρ(r) dV = ∫ρ(r) 4π r2 dr
    then putting that in your last integral and solving for f(r0) and stating f(r) = f(r0) with r0 = r.

    PS I used r0 in lieu of your R.
     
    Last edited: Sep 21, 2016
  8. Nov 22, 2016 #7
    Sorry

    What is the reason for ρ should have an infinity at r?????
     
  9. Nov 22, 2016 #8

    TSny

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    ρ is infinite at r0.

    It's because the shell of charge is modeled as having zero thickness. Since the shell contains a finite amount of charge, the density of charge must be infinite.

    It's somewhat similar to a point charge. If a finite amount of charge occupies a single point, then the density of charge must be infinite.
     
  10. Nov 22, 2016 #9
    Do you mean the volume is zero So the Q/volume=infinity?
     
  11. Nov 22, 2016 #10

    TSny

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    Yes, the volume is zero. So the volume charge density ρ is infinite. The surface charge density is finite.
     
  12. Nov 22, 2016 #11
    i See thank you
     
  13. Nov 22, 2016 #12

    TSny

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    You are welcome. I'm glad I could help.
     
  14. Nov 22, 2016 #13
    there are one more question

    How to use ρ at infinity to cancel out the dr in Q??

    can you explain more

    thank
     
  15. Nov 22, 2016 #14

    TSny

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    I'm not sure I understand the question. The idea is that ρ(r) is zero everywhere except where r = R. (I'm switching the notation from r0 to R for the radius of the spherical shell in order to conform to post # 5. Sorry for the notational confusion.) At r = R, ρ is infinite.

    So, assume that ρ(r) is proportional to δ(r - R): ρ(r) = Cδ(r - R).

    You just need to determine C. Do this by requiring ∫ρ(r)dV = Q for integration over all space. Express the integral in spherical coordinates and work out the value of C.
     
  16. Nov 22, 2016 #15
    it look likes the C should be the surface charge density
     
  17. Nov 22, 2016 #16

    TSny

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    Yes, that's right.
     
  18. Dec 2, 2016 #17
    Usually a dipole is formed as a limit as seperation d->0 but with charge q/d to compensate.
     
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